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Q.
Let $S=\{1,2,3, \ldots, 2022\}$. Then the probability, that a randomly chosen number $n$ from the set S such that $\operatorname{HCF}( n , 2022)=1$, is
Total number of elements $=2022$
$ 2022=2 \times 3 \times 337 $
$ \operatorname{HCF}( n , 2022)=1$
is feasible when the value of ' $n$ ' and 2022 has no common factor.
$A =$ Number which are divisible by 2 from $\{1,2,3 \ldots . .2022\}$
$n ( A )=1011$
$B=$ Number which are divisible by 3 by 3
from $\{1,2,3 \ldots \ldots 2022\}$
$n(B)=674$
$A \cap B=$ Number which are divisible by 6
from $\{1,2,3 \ldots \ldots . .2022\}$
$6,12,18 \ldots \ldots . ., 2022$
$337= n ( A \cap B )$
$n ( A \cup B )= n ( A )+ n ( B )- n ( A \cap B )$
$=1011+674-337$
$=1348$
$C =$ Number which divisible by 337 from $\{1, \ldots \ldots . .1022\}$
Total elements which are divisible by 2 or 3 or 337
$=1348+2=1350$
Favourable cases $=$ Element which are neither divisible by 2,3 or 337
$ =2022-1350$
$ =672$
Required probability $=\frac{672}{2022}=\frac{112}{337}$
